Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}2x-8y &= -8 \\ 2x+4y &= -5\end{align*}$
Answer: Begin by moving the $y$ -term in the second equation to the right side of the equation. $2x = -4y-5$ Divide both sides by $2$ to isolate $x$ $x = {-2y - \dfrac{5}{2}}$ Substitute this expression for $x$ in the first equation. $2({-2y - \dfrac{5}{2}}) - 8y = -8$ $-4y - 5 - 8y = -8$ Simplify by combining terms, then solve for $y$ $-12y - 5 = -8$ $-12y = -3$ $y = \dfrac{1}{4}$ Substitute $\dfrac{1}{4}$ for $y$ in the top equation. $2x-8( \dfrac{1}{4}) = -8$ $2x-2 = -8$ $2x = -6$ $x = -3$ The solution is $\enspace x = -3, \enspace y = \dfrac{1}{4}$.